In this symmetric deal, nine tricks can be made in notrump by any declarer.

Given the symmetry of the deal, we only need to analyze one declarer, so we pick South as declarer just for consistency.

Fundamentally, the problem for the defense is that even though they can set up either diamonds or clubs with one lead of the suit, both suits are blocked, and there is no immediate entry to the hand that is set up.

Isn't declarer similarly blocked? Yes. But look what happens to the deal when West leads a low club. North pitches a spade, and South wins the trick in his hand, then leads a low spade to the ten, East winning the jack, leading to: East/West's clubs are blocked, and West at the moment has no entry. What does East lead here? When North/South get back in, they have seven spades and the ♥ A, along with the ♣ K from the first trick. But at this point the defense can only take two clubs and the ♦ A before surrendering the lead.

But what if West led a diamond? Then North covers, and, whatever East does, South pitches a heart. Now, South only needs to lose one heart to take seven heart tricks, plus a diamond and a spade.

Finally, if West leads a heart, he lets East pitch a club to begin an unblock, but he does nothing to set up any of his own suits. Instead, he has blown his heart stop. Declarer just wins three hearts (the hearts are still blocked) and plays a low spade to the ten, and East is forced to win, leading to this position: Whatever East/West pitch on the three hearts and a spade, all they can take when East gets in is their diamond and club aces, and then when declarer gets back in, he takes the ace and queen of spade and five hearts, along with the first three hearts [if the defense takes the ♣ A, North must pitch a heart, but then South's ♣ K becomes good.]

Each side has two suits they might try to set up. In order to set up and run a suit, they need to lose a trick in the suit, and then either pitch a card from the Q-10-9 holding to unblock the suit, or lose the lead again to set up an entry to the long suit to make up for the blockage.

Now, when West leads a club, he does the first step - losing the mandatory club trick. But the club also lets North pitch a spade, so both sides take a step towards their respective goals. When North leads a spade to lose to East, West gets to pitch, but he holds the Q-10-9 in diamonds. If he pitches a diamond, East/West have taken one step towards establishing diamonds and one step towards establishing clubs, while North/South have taken the two steps needed to establish spades.

I have been informed that this deal was previously discovered by John Beasley, and reported in "The Games and Puzzles Journal" in 1988.

I have computed double-dummy results for all 16,777,216 different symmetric deals. "representatives." Of these deals, this is the only one where each side can make nine tricks in notrump. In fact, it is the only one that can make more than seven tricks in notrump. Is this deal unique? I can't claim that at this moment, but I would not be surprised if it was. It is certainly unique amongst symmetric deals.

There are known cases where both sides can make 13 tricks in notrump. For a simple example: Here, 7 NT makes when declared by North and East.

Richard Pavlicek reports this deal where both sides can make 9 tricks in the same suit, and it is the last making contract for both sides: Both East and South can make 3 ♠, and no higher contract makes.